Guontey:
They are actually pretty easy questions which just needs the application of Kirchhoff's law

Mbadiwe
Azikiwe
Orakiwe

I sight you sir.

The questions are quite cheap though. He is just catching cruise

thats not a tough question. the first one, apply a test voltage and measure the currents through all the branches. form a matrices and the equivalent resistance is the voltage/Current. will make use of inverse matrices.

This one dey catch cruise true true.

nextstep:
Funny... the second one is actually trival; just have to redraw the diagram a couple of times, get the parallel and equivalent resistors.
I took one look, and since I'm not busy, solved it. The voltage is approximately 0.64V.
I hope any EE students will take this serious and not let some quack lecturer intimidate you. If any lecturer says this is hard, they are either joking or not bright.

Resistance in the entire network is 10 + (2 // (1 + (2 // 10)))
In other words, 10ohms in series with (2ohms in parallel with (1 in series with (2 parallel with 10)))
then you reduce:
10 + (2 // (1 + 1.667)
10 + ( 2 // 2.667))
10 + 1.142
11.142

Observe voltage drop across the first parallel network (cd) is going to be V * (1.142 / 11.142). Call this Vcd
Then observe voltage across the second parallel network (ab) is going to be: Vab = Vcd * (1.667 / 2.667)

So: Vab = 10V * (1.142 / 11.142) * (1.667 / 2.667) = 0.64V

NOTES:
- I'm using // to mean in parallel
- The combined.resistance of two parallel resistors is well known: (R1*R2)/(R1+R2) or (2*10)/(2+10) = 1.667ohms.
- Parallel networks always have the same voltage across the networks.


I'll leave the rest of the math to the curious.

Alaye! your circuit wrong. The way to solve the second problem is to either transform form star to delta or use Maxwell's loop analysis.

drizslim:
Me thinking FG should create a separate 2 days every semester so students & lecturers get to fight

widows for Dem plenty. I swear I fit kill three

SPLENDID25:
Alaye! your circuit wrong. The way to solve the second problem is to either transform form star to delta or use Maxwell's loop analysis.

Dude what? Look at the the original circuit... it's quite simple, no need to bring in the big guns like star to delta:
- there's a 10 ohm in series with the rest of the network.
- in this network, there's a 2ohm that goes directly to ground, in parallel with the rest of the network
- in the smaller network, there's a 1ohm in series with a subset
- in the subset there's a 2ohm in parallel with a 10ohm, both going to ground.

From here it's the first week of circuits: reduce each network to its equivalent resistance, figure out the voltage drop as you go.
How is this wrong?

Intuitively, you know that 2//10 would be less than 2, and that the first series 10ohm will drop the majority of the voltage.

Na land wey dem write "military zone keep off" I wan dey sell now. Since we all wants to be mad.

nextstep:


Dude what? Look at the the original circuit... it's quite simple, no need to bring in the big guns like star to delta:
- there's a 10 ohm in series with the rest of the network.
- in this network, there's a 2ohm that goes directly to ground, in parallel with the rest of the network
- in the smaller network, there's a 1ohm in series with a subset
- in the subset there's a 2ohm in parallel with a 10ohm, both going to ground.

From here it's the first week of circuits: reduce each network to its equivalent resistance, figure out the voltage drop as you go.
How is this wrong?

Intuitively, you know that 2//10 would be less than 2, and that the first series 10ohm will drop the majority of the voltage.

this is what the circuit looks like

SPLENDID25:
this is what the circuit looks like

Good, now keep rearranging and you'll likely arrive at my diagram:

Bring in the 2ohm that's currently all the way to the right so it's next after the first 10ohm; then pull the point where the 1ohm, 2ohm, and 10ohm meet to the top right of that square; finally pull the bottom of that diagonal 2ohm to the right. You can almost visualize it.

nextstep:


Good, now keep rearranging and you'll likely arrive at my diagram.

oh,Kay! D and B are common terminals, you're right.

SPLENDID25:
oh,Kay! D and B are common terminals, you're right.

Glad you see it. Perhaps double check the math...?

And you can do the same with the first circuit too. Starting from the right... I can't see the numbers too clearly, but you can visualize:

((((10 + 2) // 6) + 5) // 6) + 4) // 8 ) + 6 and so on

then keep reducing till you get Req. The rest of the question is plug and chug. I knew some students in first year EE Ilorin who could finish this exam in under 5 minutes.

What's truly sad is that all these brilliant minds have to deal with lousy lecturers who will bill them, or be frustrated into being cab drivers and akara fryers, are persecuted by police for being smart, and in some cases killed, and often just leave the country as soon as they can. Makes me shake my head to think of how Nigeria squanders its youth.

Matrix to the rescue.
Thevenin and Kirchhoff join.
Casio calculator on the cruise.

Let the lecture catch his cruise

Bbqekpa:
Those are very simple questions na
He must be a 2.2 fresh GA with mad connections.

Smiles! For the fact that he ended the question with that hilarious crack, doesn’t mean in reality he can’t solve it. You should know that not much persons are normal in this country anymore, everything don turn cruise ���

nextstep:
Funny... the second one is actually trival; just have to redraw the diagram a couple of times, get the parallel and equivalent resistors.
I took one look, and since I'm not busy, solved it. The voltage is approximately 0.64V.
I hope any EE students will take this serious and not let some quack lecturer intimidate you. If any lecturer says this is hard, they are either joking or not bright.

Resistance in the entire network is 10 + (2 // (1 + (2 // 10)))
In other words, 10ohms in series with (2ohms in parallel with (1 in series with (2 parallel with 10)))
then you reduce:
10 + (2 // (1 + 1.667)
10 + ( 2 // 2.667))
10 + 1.142
11.142

Observe voltage drop across the first parallel network (cd) is going to be V * (1.142 / 11.142). Call this Vcd
Given two resistor networks (10 ohm, and the parallel combo 1.142ohm), the voltage over the one you want is simply the ratio of the resistor you want the drop over, divided by their sum: 1.142 / (10 + 1.142).
Then observe voltage across the second parallel network (ab) is going to be: Vab = Vcd * (1.667 / 2.667)

So: Vab = 10V * (1.142 / 11.142) * (1.667 / 2.667) = 0.64V

NOTES:
- I'm using // to mean in parallel
- The combined.resistance of two parallel resistors is well known: (R1*R2)/(R1+R2) or (2*10)/(2+10) = 1.667ohms.
- Parallel networks always have the same voltage across the networks.


I'll leave the rest of the math to the curious.

Total current=V/Rtotal= 10/11.142=0.897A
2ohm resistor between cd is parallel to 1ohm + 1.667ohms.
Hence, current through 2ohm resistor is = (0.897/(2+1+1.667))*2.667=0.574A(using current division). Hence, voltage drop across 2ohm Vcd = 2*0.574=1.148V.
Since current across Rab = 0.574, current flowing through the 1 ohm resistor will be 0.897-0.574=0.323A.
Using current division again, Current flowing through the 10ohm resistor = (0.323/12)*2=0.054A. Hence, voltage drop across Vab= 10*0.054=0.54V.

This is Kirchoffs law na. These ones never jam Mowete for unilag elect elect. You go just dey see your F for the paper.

drizslim:
Me thinking FG should create a separate 2 days every semester so students & lecturers get to fight

Like serious fist fight

A secondary school student who is good in his Physics can solve that problem.

Nothing special...

On top simple questions?

Bbqekpa:
Those are very simple questions na
He must be a 2.2 fresh GA with mad connections.

No idea what's this dude is saying here but whatever it means. he's correct

Bbqekpa:
Those are very simple questions na
He must be a 2.2 fresh GA with mad connections.

You people based everything on grades and you have 2-1 students from your place who paid their way to achieve that, can you compare them to a true second-class lower student in terms of merit?

Another point is, the lecturer must have gotten wind of what students are saying behind his back, so he decided to use their own statement against them. He won't do this without the school knowing why he did that.

Inside life..Omo some lecturers sef...wetin be d gain to spoil students result see d way labour market d hot now and graduates carry red eye d fain job..Make him and d rest Dem stay their lane oo,Biko

You call that a difficult question

Level of education sha.

drizslim:
Na questions wey I no fit solve I wan dey set” – Lecturer writes on test paper as he shocks Usman Danfodio varsity students with ‘difficult’ questions

A lecturer at Usman Danfodio University, Sokoto state has used his students to catch cruise by recoining a popular catch phrase during a class test.

The Department of Electrical and Electronics Engineering lecturer set a test question said to be difficult and hinted that moving forward, he would be setting questions he cannot solve.

In the course with code EEC211 for 200 level students in the Faculty of Engineering and Environmental Design, the instructions were for them to answer all questions.

He gave them two questions to solve and interestingly ended it with ”Na questions wey I no fit solve I wan dey set now. Goodluck”

na slaps wey him no go fit return he wan dey take now.

Aharpride1:
Comrade calm down.. 200l, doing engineering courses for the first time, 200l is the hardest in engineering for students

Indeed it is. I had my poorest cgpa in 200l

Bbqekpa:
Those are very simple questions na
He must be a 2.2 fresh GA with mad connections.

he's just playing mind games with the students..,that statement alone will make lots of people fail because after solving and getting the correct answer at ease/in time, you will doubt your answer considering the statement the lecture wrote at the end of the question and some may want to solve it another way and finally fail.

faithfull18:

I reject it in Jesus name, kilode na, I offend you before

lol...ah

I support You..

drizslim:
Me thinking FG should create a separate 2 days every semester so students & lecturers get to fight

If am not wrong you can solve the question with Kirchoff law

Bbqekpa:
Those are very simple questions na
He must be a 2.2 fresh GA with mad connections.

I dey tell you. Very simple questions.

I swear ...those lecturers are something else

drizslim:
Me thinking FG should create a separate 2 days every semester so students & lecturers get to fight